1. Add an ordered pair to the table so that the representation is NOT a function. Briefly explain why the

How do you solve for u?

Paraphin [41]

Answer:

u=9.8

Step-by-step explanation:

the minus sign on u and the minus sign on 9.8 cancel out so you get u=9.8.

3 0

3 years ago

The probability of choosing a 6 at random from a standard deck of playing cards is 1/13 what is the complement of the event of c

tekilochka [14]

To find the complement of an event you have to figure out what the probability is of not choosing a 6
since 1/13 is the probability that 6 is going to be choosen.
the other 12/13 is used for every other card accept for the 6:)
hope this is right
have a good one:)

merry Christmas:)

5 0

3 years ago

Which fractions are equievalent? a.20/5 b.-20/5 c.20/-5 d. -20/-5 e.- 20/5

pogonyaev

Answer:

D) -20/-5 because a negative divided by a negative is a positive.

Step-by-step explanation:

6 0

3 years ago

Solve 2x + 5y = –13. 3x – 4y = –8

disa [49]

If you would like to solve 2x + 5y = - 13 and 3x - 4y = -8, you can do this using the following steps:

2x + 5y = -13 /*4
3x - 4y = -8 /*5
_________________
8x + 20y = -52
15x - 20y = -40
_________________
8x + 15x + 20y - 20y = -52 - 40
23x = -92 /23
x = -92 / 23
x = -4

2x + 5y = -13
2 * (-4) + 5y = -13
-8 + 5y = -13
5y = -13 + 8
5y = -5
y = -1

(x, y) = (-4, -1)

The correct result would be D.) (-4, -1).

8 0

3 years ago

Read 2 more answers

A random sample of 49 measurements from one population had a sample mean of 15, with sample standard deviation 5. An independent

kotykmax [81]

Answer:

Comparing the p value with the significance level $\alpha=0.01$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and the difference between the two groups is significantly different.

Step-by-step explanation:

$\bar X_{1}=15$ represent the mean for sample 1

$\bar X_{2}=18$ represent the mean for sample 2

$s_{1}=5$ represent the sample standard deviation for 1

$s_{f}=6$ represent the sample standard deviation for 2

$n_{1}=49$ sample size for the group 2

$n_{2}=64$ sample size for the group 2

$\alpha=0.01$ Significance level provided

t would represent the statistic (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the difference in the population means, the system of hypothesis would be:

Null hypothesis: $\mu_{1}-\mu_{2}=0$

Alternative hypothesis: $\mu_{1} - \mu_{2}\neq 0$

We don't have the population standard deviation's, so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}}$ (1)

And the degrees of freedom are given by $df=n_1 +n_2 -2=49+64-2=111$

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

With the info given we can replace in formula (1) like this:

$t=\frac{(15-18)-0}{\sqrt{\frac{5^2}{49}+\frac{6^2}{64}}}}=-2.897$

P value

Since is a bilateral test the p value would be:

$p_v =2*P(t_{111}$

Comparing the p value with the significance level $\alpha=0.01$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and the difference between the two groups is significantly different.

7 0

3 years ago